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Thread: Cauchy Euler Differential equations

  1. #1 Cauchy Euler Differential equations 
    Forum Freshman
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    Sep 2012
    Derby, UK

    I am currently learning to solve Cauchy Euler second order differential equations, by changing the variable coefficients into constant coefficients, and then solving using the constant coefficient method.

    A question I am working on concerns the homogeneous second order differential equation:

    3x^2 y''(x) - 2x y'(x) - 2y = 0

    The solutions says that if y =x^lambda then I can work out that y'' = lambda x ^(lambda-1) and y'' = lambda(lambda-1)x^(lambda-2).
    All fine so far.

    However it then says that if I substitute y= x^ lambda into the given differential equation I will get:

    (3 lambda (lambda-1) x^lambda) - (2lambda x lambda) - 2x lambda = 0

    I was just wondering how this has worked out?

    Any pointers would be great. As I have substituted y= x^ lambda into the given differential equation and not got anything close except for the last term.

    Kind regards


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  3. #2  
    Forum Freshman
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    Sep 2012
    Derby, UK
    Never mind. I worked it out eventually with lots of algebra. The book I have is terrible at explanations!

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